Does validity of contrapostive proofs require the Law of Excluded Middle?

Question

I remember that during my first proofs class the hardest thing I had accepting were the logic we had to learn, and it seems I still have questions about.

So I was thinking about why when the contrapositive is proved true then it implies that the original statement was true. The way I've been thinking about it is by considering a statement about an element of some set $A$. Letting $Q(x)$ represent $x\in A$ such that this statement holds true for, so the way I've translated $$Q\rightarrow P\Longleftrightarrow \sim P\rightarrow \sim Q$$ to $$Q(x)\subseteq P(x)\Longleftrightarrow P(x)^{c}\subseteq Q(x)^{c}$$ This makes sense initially to me since it does seem that elements that follow $\sim P$ would be element that don't belong to $P(x)$ (i.e. they belong to $P^{c}(x)$) Thus this would make sense to me because $P(x)\cup P(x)^{c}=A$ since it seems that either P or $\sim P$ must hold for an element.

I guess the main question if this last statement is possible would rely on: is the law of excluded middle always hold? Could you perhaps have a nonsense statement, so its negation is also nonsense, and no possible element is from either. Or perhaps the way I'm thinking about contrapositive statements is wrong?

Answer 1

In classical logic every statement has a truth value, even if it is nonsense like "if the Moon is made of cheese then the sky is made of rubber" (this one is true because any implication with a false premise is defined to be true). The law of excluded middle is a law of classical logic, so it is always true as an axiom, in particular for any set either $x\in A$ or $x\notin A$ is always true.

There are alternative systems of logic where this law is not adopted, intuitionistic logic for example, but there the interpretation is not that $x\in A$ and $x\notin A$ are both nonsense, but rather that there is no "constructive" way to establish either. See more in Can one prove by contraposition in intuitionistic logic?